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Using the extended fundamental counting principle, you multiply the total number of options in each space together. There are 10 possible numbers for each of the three number spots, so you would do 10x10x10=1,000. Multiply this by 26 and 26 again for all the possible letters that can go in each letter spot:

1,000x26x26=676,000

So you have 676,000 possible license plate combinations.

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โˆ™ 2010-03-18 01:20:11
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A polynomial of degree zero is a constant term

The grouping method of factoring can still be used when only some of the terms share a common factor A True B False

The sum or difference of p and q is the of the x-term in the trinomial

A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials

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Q: How many combinations are there for a license plate that has 3 numbers from 0 -9 and 2 letters from A-Z?
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How many possible combinations could be constructed using 3 letters and 3 numbers for a license plate?

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